Due on Feb. 2nd at 4:30 pm. Please turn in a physical copy unless you are absent, in which case you may send me an electronic copy.
For a volume charge density , the electric field at is given by
where and is the location of a charge.
For a point charge at the origin, . Evaluate the above integral using cartesian coordinates and unit vectors and this .
For a point charge at , . Evaluate the above integral using cartesian coordinates and unit vectors and this .
Solution
1.
The generalization to the 3-D problem given is straightforward.
2. Here , and are fixed points in space. Normally when given a charge density, we replace with . If we did that here, we would have , which does not depend on primed coordinates. To work around this issue, change the integration variables to be double primes. As before, in two dimensions,
In the integral above, the is a constant with respect to integration (in the same way that in , is a constant) and so we replace with this constant value.
In class, I used an alternative approach where I replaced with and integrated over so that I needed to evaluate
The result of integration is
To finish the problem, I replaced with the variables given in the problem of to get the same answer as when integration was done over double primed variables.
Suppose . Find and in terms of . Show graphically why the result makes sense using .
Show that . (That is, show that ).
Solution
1.
2. The generalization of the previous result is if , then for
.
In the case that and a finite integral, it follows that , which applies to and thus whe have shown .
Griffiths 4th Edition example 1.15 gives an alternative solution. Another approach is to think of this as a transformation of variables problem in which the Jacobian, which involves the absolute value, is computed.
Find using either
identity 5. on page 26 of Jackson, 3rd Edition or
using the substition .
Prove identity 5. on Page 26 of Jackson 3rd edition
where is assumed to have only simple zeros, located at . Do this by either
explaing the omitted explanations in Dennery and Krzywicki 1967, page 237 or
Taylor series expanding around the points where .
Solution
1.1 To avoid confusion with the in the given equation, re-write the identity as
Here , which has a single zero at . , so
and so
1.2 With the substitution,
for positive and for negative , so the integral is as before.
Find the electric field due to a non-conducting sphere with a uniform volume charge density and radius that is centered on the origin.
Show by direct calculation that
when the volume of integration is a sphere centered on the origin with
radius
radius
Solution
As mentioned in class, this is somewhat of a simple problem but was given to motivate the use of the Dirac delta function. What I said in class is briefly summarized at the end.
1. The electric field is
where was used.
2. In the following,
is used. This follows from the identity for the divergence in spherical coordinates.
For and a spherical surface of radius :
The divergence of is, using ,
as expected from Gauss’ law in differential form ( with ). Therefore, for :
Conclusion: for ,
For and a spherical surface of radius :
The divergence is
also as expected from Gauss’ law in differential form ( with ). Important: the last equality in the above equation is only true when ; when , the result is . This issue is considered later in this solution.
One must split the volume integral into two integrals, one for and the other for to account for the divergence being different in the two regions. As shown earlier, in the region , the divergence is zero.
Conclusion: for ,
Motivation for this problem
I gave this problem as a lead-in/follow-up for my discussion of the motivation for the delta function. For a point charge, the integrand of
with
is zero everywhere except at the origin () where it is indeterminate in the form of because
However,
does not have an indeterminacy. Therefore, the divergence theorem
applied to for a point charge gives
In this problem, it has been shown that if a point charge is modeled as having a uniform density for and having zero density for , then the indeterminacy problem can be avoided and
for all and for arbitrarily small (but non-zero) (but see also ‘other notes’ below). That is, the divergence theorem gives the correct answer for a solid ball of charge that is arbitrarily small but not zero.
Although the divergence theorem does not apply for , to avoid having to think about a solid ball with arbitrarily small but non zero radius is to use it with the convention that
where the function is zero everywhere except at and integrates to 1 when the volume of integration includes .
That is,
When you see the delta function used in an integration, you should think back to this problem and realize that the delta function is being used instead of modeling the electric field as a point charge as having a uniform density for and zero density for .
Griffiths takes a different approach to motivating the introduction of the delta function. He notes that the indeterminate integral of the divergence
must be equal to because of the divergence theorem (which technically does not apply because of the singularity in at ):
He then notes that has the property that its divergence
is zero everywhere except at the origin (where it is interminate)
integrates to a constant () over any volume that includes (based on the assumption that the divergence theorem applies and resolves the indeterminacy)
That is, although
has an indeterminate integrand at , the divergence theorem must be true, and so this integral must be equal to the surface integral, which is . Therefore,
where the function is defined to be zero unless and integrates over a volume that includes to 1. This explanation is somewhat awkward because it asserts that the divergence theorem must be correct even though the field does not satisfy the requirements for the divergence theorem to apply. (The requirements for the divergence theorem to apply are that the components of the vector field and their derivatives are continuous.)
In summary, the integral
is indeterminant when and includes . If we model a point charge as having a uniform density for , and zero density for , where is smaller than any length scale in the problem, the indeterminacy is removed. Instead of two integrations, one for and the other for , we can simply replace with ; the result of any integration will be the same.
A second motivation for using the delta function is that we often want to do an integral of the form
(Recall that means ). As before, integral is indeterminate because is 0/0 at . To work around this problem, we can again model the point charge as having a uniform density for and zero density for . Then
The second integral is identically zero because for . If can be expanded as a Taylor series
then
The second and higher-order terms in the last equation approach zero as . To see this, note that for small enough ,
where the is the average in in a sphere of radius . Therefore,
The same argument applies to the higher-order terms in the Taylor series expansion.
Prior to the introduction of the delta function by Dirac in the 1930s, the above argument was used when an integral of the form
was encountered and was that for a point charge and that included the location of the point charge.
With the introduction of the Dirac delta, one can simply use to get the same result without needing to think about the above steps:
That is, using the delta function for the divergence of due to a point charge will give us the same answer as if we had modeled the point charge as having a uniform density for , zero density for , Taylor series expanded , and then let .
Other notes
In the above, I showed that the divergence theorem gave the correct result for and any by direct calculation. However, the divergence theorem
requires that within , the components of and their derivatives are continuous. In , is continuous, but is not. So it appears that direct calculation for gave a result that was consistent with the divergence theorem even though the assumptions required to apply the divergence theorem are not satisfied. To see why this occurred, use the divergence theorem in the two parts of where the divergence theorem applies for .
First, consider the application of the divergence theorem for the volume between and .
The outward normal to for is , so
Next, consider the application of the divergence theorem for the volume between and (the result will apply to any ). In this case, there is an inner and outer surface and so the surface integral has two parts.
The outward normal to for this volume is and the outward normal to is , so
Using and with the volume integral expressed in two parts
gives
(The surface integrals at cancel because their normal directions are in opposite directions.)
You do not need to turn these problems in. If you turn in these problems, I’ll provide feedback.
These are problems that either I discussed in class or are related to the topic covered in class. I place them here for your reference. Some of the problems listed may not have been covered in class but may serve as exam practice problems.
A straight 1-D rod of mass with a uniform mass density and length () is aligned with the –axis and centered on . Find the center of mass by evaluating
Find the moment of inertia about the –axis of the rod in the previous problem in terms of , , and by evaluating
Explain why the value of for makes sense.
A spherical shell has a net charge uniformly distributed on its surface.
Find in terms of the delta function.
Show by direct calculation that
is satisfied when and is an origin–centered cube of side length with its sides parallel to the coordinate planes.
The step (or Heavyside step) function, , is defined by
can be used to to express a charge density in a compact mathematical form. For example, instead of stating “a uniformly charged sphere centered on the origin, with total charge , and radius ”, we can write . When this density is integrated over all space, the result is :
Because for and for , the limits on the integral can be modfied and the function removed:
The integrals above evaluate to the volume of a sphere of radius so that
Write the piecewise function using one or more functions.
Find the volume charge density in cylindrical coordinates in terms of and/or for an infinitely long cylinder of radius with a charge density per unit length of uniformly distributed on its surface. Assume that the cylinder’s centerline is along the -axis.
Repeat 2. assuming the cylinder is finite and extends from to .
Solution
As mentioned in class, an easy way to solve this is to write as the sum of two step functions. One shifted to the left by , corresponding to ; the other shifted to the right and inverted, corresponding to . Thus,
.
Other solutions:
It may be interesting to use develop an identity for similar to identity 5. on page 26 of Jackson 4th Edition.
Checks:
when ,
when ,
when ,
We want a charge distribution that is localized at a cylindrical radial distance . has this feature, where is a constant. We also need the integral of this charge density over all space to equal the total charge on the cylinder, which is assuming that the length of the cylinder is . Thus, we need to find such that
and so
Equivalently, we can write
because , which follows from the fact that the delta function is only non-zero at .
Here we apply the result from part 1. with the replacement of in that problem with :
For discrete charges, Green’s Reciprocity Theorem for charges is
Consider charges , and at locations , and , respectively, on the -axis. At these locations, the potentials are and , respectively.
If a new system of charges , and is created by placing them at the same locations , and , respectively, the potentials at these locations are and , respectively.
Show that
Solution
The physical interpretation of Green’s Reciprocity theorem is that the work required to put charges an unprimed set of charges at under the influence of only the electric field of a primed set of charges must be the same as the work required to put a primed set of charges at under the influence of only the electric field of an unprimed set of charges at .
Using
gives, for , the left-hand side terms in
The right-hand side terms in can be expanded as
In the above, the left– and right–hand sides of the equation to prove were written in the form of a matrix. In this form, one can see that the matrices are transposes of each other. If we sum all elements in a matrix, the result is the same if we sum all elements of that matrix transposed.
More generally, one can use
Repeating with the primed and unprimed swapped and starting with a dummy index of instead of , we have
Where in the last step, was used.
One can think of as elements in a matrix. The equations
and
both result in the sum of all elements of a matrix excluding the diagonals. The only difference is the order in which the elements of the matrix are summed. For a continuous charge distribution, we would use the fact that the order of integration could be changed to arrive at the result .
In class, I partially did problem 1.13 of Jackson 3rd Edition (I only found the net charge on the upper of the plate). Find the net charge on the lower plate. Justify your steps at the level of detail given in class. (I stated an easy way of finding the net charge induced on the lower plate, but I want you to do it the long way, which requires steps similar to the ones used in class.)
A point charge at a distance from the origin is between two grounded spherical conducting shells of radius and that are centered on the origin. Find the net charge induced on the surfaces at and .
For discussion during the next class: In my solution to part 1., I used the continous form of reciprocity. Could I have solved it using the discrete form?
Solution
1. Let the unprimed system have a conducting plate in plane held at and a conducting plate in the plane held at at .
Let the unprimed system have a conducting plate in the plane held at at , a conducting plate in the plane held at , and a point charge at .
One can use the equation from problem 1.11 of Jackson or start with
and split the charge density into surface and volume charge densities and , where the subscript means charge not on the conducting surfaces :
The first term on RHS is zero because (there are no charges in the unprimed system between the plates). The second term on the RHS is zero becuase on conducting surfaces.
The first term on the LHS reduces to becuase (the notation is also acceptable; this gives , which is the same things as notationally). The potential between the plates in the unprimed system is
and so the first term on the LHS is .
The second term on LHS is composed of integrals over the top and bottom surfaces (assuming all of the charges in the primed system are on the plates or at ; this is discussed below). Given that on bottom surface, we are left with an integral over upper surface () for which .
The equation
thus reduces to
One can reverse the potentials on the unprimed system to arrive at the charge on the lower primed surface, which is
Note also that the assumption that the total charge in the universe is zero
could have also been used to find given .
Note: In the solution, I assumed all free charges were on the upper and lower surfaces and there were no charges on far-away surfaces; this statement is justified post hoc by uniqueness and because computing and in this way gave . To avoid this assumption, one would need to use equation 1.35 of Jackson with the volume being the volume between the plates. Curiously, in problem 1.11, Jackson suggests using equation 1.35 to come up with what he calls Green’s reciprocation theorem, seemingly so it could be applied in problem 1.12. However, the equation in problem 1.11 does not directly apply because the surface of the volume between the plates is not all conductor (only the top and bottom parts of the volume are conductors, the sides of this volume are not). One is left with having to use equation 1.35 and the argument that because the problem is 1-dimensional, the derivatives of the potential with respect to the normal on the side surfaces of the volume are zero.
2. The procedure here is nearly identical to part 1. The potential between the sphere in the unprimed system when the potential at is and the potential at is zero is
which can be found by solving the boundary value problem
to find and or by assuming a charge on the outer sphere and on the inner sphere and using Gauss’ law to find and then (the will cancel). The final result is
In one of the steps required to derive his identities (discussed in 1.8 of Jackson, 3rd Edition), Green said: “Consider the divergence of a scalar function multiplied by a vector function ”.
Show that
(this is equation 1.32 of Jackson)
Show that
where is the volume enclosed by the surface , is the normal to a differential area element on , and is a differential volume element. (This is needed in the following problem.)
Verify the result in 2. by using an , and of your choosing.
Solution
1. This can be shown by writing and in cartesian coordinates.
2. The divergence theorem
using on the LHS and the the RHS gives
and rearrangement gives
Note that when , , and is a cube, this reduces to the integration by parts formula.
Consider the integral
Use and the result derived in the previous problem to re-write this equation in terms of a volume integral + a surface integral.
Repeat 1. using
Show how the results from 1. and 2. can be used to show
which was a result derived in class last week using the discrete form of Green’s reciprocity theorem.
Solution
The identity from the previous problem is
or, switching locations of the volume integrals,
Using and in the identity and gives
Using and in the identity and gives
(The two equations for must be equal because the integrands on the left–hand side integrals are the same because .)
Equating the two equations for gives
If the volume is all space, the area of becomes infinite. If all of the charges that make up and are confined to be in a sphere centered at the origin with a finite radius, then as the radius of , and and similar for and . (You needed to provide some sort of justification similar to this in you solution - saying the potential and field approaches zero is not sufficient because the surface becomes infinite and you are left with ). So for being a spherical surface with large
Where and are the total charges in the unprimed and primed system respectively. As a result of this argument, the surface integrals can be dropped and can be replaced with and we are left with
Finally, using and gives
You do not need to turn these problems in. If you turn in these problems, I’ll provide feedback.
These are problems that either I discussed in class or are related to the topic covered in class. I place them here for your reference. Some of the problems listed may not have been covered in class but may serve as exam practice problems.
Find , where is the electric field due to a point charge at and is a sphere centered on the origin with radius .
A point charge is at a distance from the center of a conducting sphere of radius . Use reciprocity to find the potential on the surface of the sphere.
Find the volume charge density in cylindrical coordinates in terms of and/or for a uniformly charged disk of radius with charge density that lies in the plane and is centered on the origin. Use for the radial coordinate in cylindrical coordinates.
Use identity 5. on page 26 of Jackson to convert this to spherical coordinates. Use for the radial coordinate in spherical coordinates.
Two infinite and grounded conducting sheets are in the and plane. In the plane, there is an infinite non-conducting sheet with surface charge density .
Find the potential, , on the left () and to the right (), , of the non-conducting sheet using any method (Gauss’s law or the boundary value method can be used; you should be able to do it using both methods, but you need to only show your work using one method).
Write the potential for as a single function using and and the Heavyside step function . (In the future this (with set to 1), will be called a Green function, which is the motivation for the title of this problem.)
Show that . You will need to use the fact that , and . Also compute , where the prime means to take derivatives with respect to primed variables. (This may seem odd because was defined to be a constant; here you are being asked to treat it as a variable. You should get an answer that is proportional to ).
As discussed in class, the motivation for solving this problem is that its potential, , can be used in Green’s second identity (eqation 1.35), which is a form of reciprocity, to solve the most general problem for this geometry. The most general problem is to find the potential when between and , the left plane is grounded, and the right plane is at potential . Instead of using found above to solve the most general problem, first use it to solve an easier problem:
Use Equation 1.35 and to find the potential when between the conductors and and . Include a sketch or a sentence where you define and when you use Equation 1.35. There will be a subtilty with notation here – if you use Equation 1.35 as written, you’ll end up with and not the desired .
Solution
1. One can answer this question by assuming a surface charge of and is induced on the plates at and . Then use the formula for the electric field of the sheets of charge at at , at , and at to get the total field and . The two unknowns, and , can be found by using the fact that the electric field in either of the conductors is zero - this yields , which is expected because the net charge in the universe is zero. The second equation needed to find and in terms of is
Alternatively, one can use the fact that has solutions of the form , the two boundary conditions and the continuity and jump conditions at .
Boundary conditions:
This leaves
and
Continuity:
Jump:
The jump condition follows from Gauss’s law with a Gaussian cylinder with one end cap to the left of and the other to the right. It also follows from integrating once with respect to . The condition is
or
The three equations left are
Solving gives
and
So the final solution is
Note that swapping and in gives the equation for , and vice-versa. That is, .
2.
Checks:
When , and , giving .
When , and , giving .
3.
If one plots , you will see that its derivative has a jump downwards at of and so we expect the terms in with the derivative of the Heavyside step function, which is the delta function, to be zero. So the two terms involving derivatives of will be addressed first.
The first term can be rewritten using
The first equality follows from the identity given in the problem statement: . The second equality is a standard delta function identity.
The second term is
The following two equalities follow from the fact that is zero except at :
Given that , the two terms in involving derivatives of cancel, leaving
Straightforward calculation and use of the same identities as above gives
This matches expectations. For a sheet of charge in the plane, the volume charge density is
and Poisson’s equation is
(Recall that the units of are inverse length. This follows from part of the definition of the delta function: ).
4. Equation 1.35 with replaced with is
Let be the volume spanned by , , and , where is arbitrary. This volume has six surfaces and so the surface integral will have six parts.
In the volume integral, the term with because there is no charge in the volume of the system. Subtitution of gives
Integration over gives
Of the six sides of the surface, on all except the one in the plane for which the outward normal is . So
where found in part 1. was used.
One could assert that we expect to depend only on , so on the other four faces, for which . Alternatively, one can use symmetry make the same conclusion:
The sides of that are in the and plane have , so two of the surface integrals of are zero.
On the surface in the plane, , so we need to evaluate
On the surface in the plane, , so we need to evaluate
Because the geometry is invariant with respect to , the two partial derivatives are equal and the sum of these two surface integrals is zero. An idential argument can be made for the surfaces at and .
All of the parts of the volume and surface integrals have not been evaluated and so equation 1.35 reduces to
or
In the problem statement was requested. In the above equation, we have as a function of , which was a constant in the problem. Given that both and , we can find any value of in this range by solving a problem with set to that value. Thus, we can equivantly replace with and so we have shown
Two conducting and grounded spherical shells of radius and are centered on the origin, and .
A nonconducting spherical shell is centered on the origin and has a charge density of and radius , with .
Find , the potential between the inner conducting shell and the charged shell and , the potential between the charged shell and the outer conducting shell. (Read the subscript as “inner” and as “outer”.)
Find the surface charge densities on the inner and outer conductor.
Verify that Gauss’s law is satisfied for a Gaussian sphere centered on the origin and with a radius that is between the charged shell and the outer conducting shell.
Write the potential for as a single function using and and the Heavyside step function .
Solution
1. Instead of starting with and , we immediately write
and
because we know the potential must have a dependence and the above two equations satisfy and . What is left then is to use the conditions and to find the two unknows and . The result is
Written in terms of the total charge on the shell with charge density at , this is
As before, there is symmetry with respect to swapping the spatial coordinate with a parameter : (note that this is only after writing the potentials in terms of the charge on the shell at ).
2. From Gauss’s law, near the surface of a conductor , where is in the direction perpendicular to the conductor and outwards. For the inner conductor, ; for the outer, . (Note that in Green’s second identity, the convention is that is in the direction perpendicular to and outwards.) Evaluation givves
The charge induced on the inner and outer surface is
Checks:
1. As , and .
2. As , and .
3. In HW #2.3.2, we found the charge induced when a point charge was at . We can use that answer and superposition to find the total charge induced if point charges are distributed uniformly on a sphere. This should match the total charge induced found in this problem. From #2.3.2, the answers were (with replaced with ):
The “superposition” argument requires a bit more justification. Consider two separate problems. In one problem there is at ; call the potential for this single problem . This potential will satisfy
with at and at
In another problem, there is at ; call the potential for this single problem . This potential will satisy
with at and at
The sum satisfies
with at and at
This argument can be extended for an arbitrary number of differential charges on a sphere of radius .
We can conclude that the sum of the potentials for isolated charges at any position on a surface of radius and satsifies Poisson’s equation and the boundary conditions, which is the same condictions that we require for the combined charge problem. The charge density obeys superposition: so that the charge densities on the conductors for two charge case is the sum of the charge densities for from each individual case.
Another way of justifying superposition is to note that for the isolated problem, the net force on the charges induced on the conductors is zero. If is introduced the net force on all of the induced charges will still be zero.
Before stating Equation 1.42, Jackson (3rd edition) notes that with Equation 1.35 and 1.39, it is simple to obtain a generalization of Equation 1.36.
Show the “simple” steps need to arrive at Equation 1.42.
Be very careful with notation. You’ll need to think a bit about the justification and validity of swapping and and/or changing the dummy variable used in integration. When you do either of these, provide a justification so that I know you are not applying the “answer operator”.
Solution
In HW 3.1, the use of Green’s second identity resulted in a potential that depended on the primed variable because integration was performed over an unprimed area or volume. In that problem, an argument was given for why the prime variable could be simply replaced with the unprimed variable. In principle, one could always use the approach used in HW 3.1 to find and then make an argument for why is the equation for with the prime removed. In this problem, we follow Jackson’s logic for Equation 1.42 for which the result of integration is , which depends on the unprimed variable.
Equation 1.35 is
Using, , , and gives
We want the result after integration to depend on , so change the integration variable to be primed. The derivatives must also be changed to be primed and must be replaced with .
Equation 1.39 is
where the prime means the partial derivatives are taken with respect to primed coordinates. In the above integral, we have . For now, assume that . The first then term evaluates to
with this and rearrangement, we can write
To get equation 1.42, one must justify
For a general function, say, , . Showing in general is non-trivial and Jackson problem 1.14 gives a suggestion for the proof as does the second paragrapha of page 40. A footnote on page 40 gives a reference of the proof for 1.14(b). So ideally one would have cited the statement on page 40 of Jackson to justify the use of .
In class, I showed how to find the potential due to a point charge inside of a grounded, conducting, and closed dome when the radius of the dome is infinite. The potential was found using the method of images. (A dome is a hemisphere plus a cap.)
Use this and equation 1.35 of Jackson (or equation 1.42; is proportional to a Green function) to find the potential for the following problem:
An conducting dome has its base in the plane and the radius of the dome is infinite. There are no charges inside of the dome. The boundary of the dome is grounded except for the region between , which is held at a potential of .
Find the potential inside of the dome (the integral required to find requires an approximation of the integrand or numerical integration and will be considered in the future).
Find the potential inside of the dome when . Do this by Taylor series expanding for small . Does the answer make sense?
Sketch the electric field lines inside of the dome.
Note that a related problem (but with a different Green function) considered in 2.6 of Jackson.
Solution
1. The potential that satisfies the boundary conditions in the limit that that that the radius of the dome is much larger than is
This equation was derived using the method of images in class and is the same equation that solves the problem for a point charge at above an infinite grounded plane in the - plane. satisfies the boundary conditions that when , and when , (showing the latter requires re–writing the denominators in terms of and ).
Given that we are only interested in the potential along the –axis, we can set . (In 2.6 of Jackson, he does this after the integral required to find is found.) We then have
Green’s second identity (Equation 1.35) with in place of is
In this problem is the volume of the dome and has two surfaces – a large disk in the – plane and the curved part of the dome.
The Laplacian of can be written down by inspection. The charge density is that for point charges at , so
is zero in because and are positive in . The integral simplifies to
.
Inside , we were given that there are no charges, so . Therefore, the left-hand side of Equation 1.35 reduces to .
On , , so the second surface integral is zero. We are left with
The surface is composed of the two parts – the curved part and its base. On the curved part, was given. On the base, for and otherwise. As a result, the surface integral reduces to
.
In cylindrical coordinates, we have
The outward normal to is -, so
What remains is to evaluate the partial derivative (and recall that in the integral means ). In this problem, “” corresponds to .)
Evaluation of the integral gives
Cancelling constants and using for gives
Using the same arguments used in HW 3.1, we can replace with , leaving
2. The solution can also be written (for ) as
Using the binomial expansion,
To lowest order in ,
For large , the (positive) charge on the disk appears as a point charge at the origin. As a result, we may expect that the potential will fall off as . Here the potential falls off more slowly. The reason is that there is also a contribution from the negative charges in the plane for . As a result, for large , the system appears to have zero charge at the origin and so there is no term. (The interpretation of the power series expansion of potential will be covered in more detail later.)
3. This was discussed in class. In general, it is easiest to draw equipotential lines starting with one near the boundary. Then, draw electric field lines that are perpendicular to the equipotential lines.
Review Griffiths 3.3.1 (3rd and 4th edition). You should be able to quickly derive the potential for Figure 3.20 (3rd and 4th edition) when any one of the sides is at and the other three are grounded. See also my PHYS 305 notes.
Suppose the sides of Figure 3.20 of Griffiths 3rd or 4th edition are held at , , , and , where left, right, top, and bottom. Find the potential inside the rectangular pipe.
Hints:
The total potential will be the the sum of the potentials from four different problems (1) non–zero and all other sides zero, (2) non–zero and all other sides zero, etc. I will ask you to explain why in class.
You can build the solution to three of the four problems described in the first hint using a transformation of coordinate system and the solution to one of the problems.
Solution
Note: Diagram and solution below assumes width is and height is . Diagram in Griffiths has width of and height of .
The potential inside of the long (in direction) duct with cross-section is shown in the following figure
is
The geometry for the case is
The choice of coordinate system for was arbitrary and we can translate the duct used for to the left by and down by so that the position of the duct is the same as that in Figure 3.20 of Griffiths.
This is effected by replacing with and with in :
Check: When , and when , .
To find , replace with and with in the equation for :
Check: When , and when , .
To find , swap and , and , and with in the equation for :
Check: When , and when , .
To find , replace with , and with in the equation for :
Check: When , and when , .
The total potential can be written as
This potential satisifes Laplaces’s equation and the boundary conditions for the case where all four sides are at a different potential.
In 2.7 of Jackson 3rd edition, he solves for the potential outside of a sphere with hemispheres held at using a Green function. In example 3.7 of Griffiths 4th edition (example 3.6 in 3rd edition), he describes how to solve for the potential outside of a sphere with a potential of on its surface. As a result, Jackson’s example is related to Griffiths’ example.
Study these two examples and be prepared to answer questions (such as how they are related) about them in class.
The potential inside of the long (in direction) duct with cross-section is shown in the following figure
can be written as
In Example 3.3 of Griffiths, he finds the potential for a long duct with one open face. The diagram for that example is similar to the above except there is no conductor on the right side and . The solution he finds is
1. Show that the infinite solution () matches the finite solution () in the limit that . (You do not need to derive or , but should be able to.)
2. In the limit that , it would seem that near the center of the duct (), the system appears as two infinite parallel conducting plates held at different potentials, which is a 1-D problem solved before. Show that in the limit , equation reduces to the solution for an infinite conducting plate in the plane held at and a grounded infinite conducting plate in the plane. (You’ll need to use for multiple times and also the Gregory and Leibniz formula for ).
Solution
1. We need to show that
reduces to when , or, equivalently,
Re-writing the numerator and denominator in terms of exponentials
gives
In the numerator, the term as for .
In the denominator, the term as .
As a reuslt, the ratio approaches as expected.
2. We need to show that near , the sum in approaches so that .
In the middle of the duct, so we replace with for . Using and for gives
We are left with
From mathworld.wolfram.com,
Using , we can re-write this as
Finally, for , and so
Thus,
An infinitely long, hollow, and conducting duct is parallel to the -axis and has sides bounded by and . All sides are grounded. An infinitely long non-conducting sheet of charge with surface charge density is in the plane between and . See the left part of the figure below.
1. Find equations , the potential for , and , the potential for . Based on the discussion in class, we can easily write down equations for and that are zero on the conducting walls:
where is an integer, matches the boundary conditions on the walls of the conductor. You need to find and such that the two conditions
and
are satisfied.
2. Write a single equation for the potential using , , and the Heavyside step function .
The Green function is , where and is the length of the duct. (You may want to verify this.)
3. Use and Green’s second identity (Equation 1.35 of Jackson), or equivalently, the Green function stated above and equation 1.42 of Jackson to find the potential when the duct is filled with a nonconducting material with a uniform charge density as shown in the right part of the figure above.
Note: In problem 2.15 of Jackson, he gives the Green function for the case where a line of charge parallel to the -axis passes through . That problem is a bit more complicated than the problem considered here. In problem 2.16, of Jackson, he states the result of using the Green function from problem 2.15 to find the potential for the configuration in part 3. above.
Solution
1. Both
and
satisfy and the boundary conditions
The equations for and could have been derived using the above six boundary equations and starting with
or most generally
However, the result will be the same. In problems such as these, it is often easier to reason out what the form of the solution will be rather than starting from the most general equation. In this case, we need the solution in the direction because it can have zeros at two locations. The term is justified because we know that and can be combined into a with a “phase” shift .
The continuity condition
gives
It follows that
The jump condition
gives
The term in square brackets
rewritten using the continuity relationship is
or
The numerator simplifies to . This can be seen by expanding the hyperbolic functions as exponentials or using the identity
with and .
The continuity condition equation thus simplifies to
Multiplication of this by and integration from to and using
gives (after changing the dummy index from to ),
for and for .
or, using and defining to the the smaller of and and to the the larger of and ,
or, in terms of ,
2. Using gives
3. The surface integral in equation 1.44 of Jackson is zero (why?) leaving
does not depend on or and , so integration over and gives
where is as defined in the problem statement. Using from part 2. gives
The step function modifies the limits of integration, so
The interpretation of this equation is that to find the potential at , sum over the potential due to differential sheets at different . The first integral is the potential just to the left of all differential sheets to the right of . The second integral is the potential just to the right of all differential sheets to the left. This is the equation you would probably write down in order to compute the potential due to a slab if you did not know about Green functions – the total potential is simply the sum of potentials due to all differential sheets.
The first integral requires the integration
The second integral requires the integration
Using these, evaluation of the integrals gives
Using again
with and
To combine into one term, note that when expanded in terms of exponentials the following sum has four terms
The following product has four terms
Setting and
gives and . Thus
and
Finally, using
gives
and the final result of
Defining gives the result of problem 2.16 of Jackson.
A non-conducting and infinitely long cylindrical shell of radius has a surface charge density of .
Find using the boundary value method. Note that the general solution to Laplace’s equation in cylindrical coordinates for a problem with no dependence is
Repeat 1. assuming . In this case, the term cannot be dropped for the solution; is determined from the requirement that for large , the potential should be that for a line of charge with linear density that is the average of over .
Solution
I’ll solve part 2., which reduces to part 1. if . I will also solve this the long way. After solving this problem the long way, you should be able to identify short–cuts, one of which is find the potnential due to a long and uniformly charged cylinder and adding the result to the answer of part 1.
We can conclude that the terms are zero for the potential outside of the cylinder because at large the cylinder will be approximately that for a line of charge, for which potential will be proportional to for large , where is the azimuthally averaged surface charge density: . That is, for large , the potential should be the same as that when all of the charge on the cylinder is collapsed onto a line, for which the potential is , giving and . The constant is determined by defining a reference potential at a reference point . For example, if , then and the terms combine to form
Note For a line of charge, we can’t choose a reference potential to be zero at infinity. This is related to the problem that arises if one attempts to use to compute the potential for an infinite line of charge. The derivation of this integral assumes that the potential is zero at large . Using Gauss’ law, one can show that this is not the case – and so , where is a constant. We can’t choose because it gives . So to compute the potential for an infinite line of charge, one must either use the integral equation for the electric field or Gauss’ law. See also 2.3.4 of Griffiths where he notes that the integral diverges for a charge distribution that extends to infinity.
Outside the cylinder, the general form is then
I’ve kept the term. In the following, I’ll show how the values of and given above follow from an alternative argument than the one given above.
Inside the cylinder, the potential should be finite and so the and terms are dropped. (If the cylinder had a constant charge density, we would expect the potential to be constant inside of the cylinder and would keep only the term.)
Inside the cylinder, the general form is then
The two boundary conditions are
and
(Make sure that you know how to derive the second boundary condition.)
The boundary condition gives
This equation gives
and
from “matching terms”, which means integrating both sides by either or , with being an integer greater than zero, and integrating from to and using
Notice that in cylindrical problems we integrate over , so the limits of integration are from to . In contrast, in spherical problems with a dependence, we integrate from to .
To determine the relationship between the coefficients quoted above, it is often easier to write out the first few terms in the sums
in alignment as
and then read off the matching terms
in order to conclude the general result stated earlier of
The boundary condition
gives
This gives
The equation from the second boundary condition combines with from the first boundary condition to give
The equation from the first boundary condition combines with from the second boundary condition to give
which is only possible if . Similarily, all terms except and are zero.
The equation from the first boundary condition and combine to give
The final result is
If the reference potential is chosen to be zero at , then and
(In this problem, the reference potential was not given intentionally. Ideally you recognized that a reference potential was needed to fully specify the solution. Also note that we cannot specify a constant reference potential at any other because the potential depends on .)
Note: Inspection of this answer suggests a faster way of solving the general problem of
or more generally,
First find the potential due to . Then, write and not as an infinite sum, but rather as a sum that involves only terms of and that appear in .
Check: Earlier it was argued that for large , should approach
where
Using gives and so the computed value of of
matches the limit
Do problem HW problem 5.2.1 with the modification that and show that how it can be used to arrive at the result quoted in problem 2.15(b) of Jackson.
Use your answer to part 1. and reciprocity to find the potential inside the tube when inside the tube, the side at is held at a potential of , and the other three sides are grounded. (This is an analog to HW #3.1.4 where we solved a problem that was easy to solve without reciprocity using reciprocity).
Solution
1. The steps used to derive
and
were given in HW problem 5.2.1 and these equations are the starting point here.
For this problem, replace with in the above equation.
Multiplication of both sides by and integration from to and using
gives, after changing the dummy index from to ,
for . Note that is not constrained to be odd as it was in HW problem 5.2.1. The new result is
or, using and defining to the the smaller of and and to the the larger of and ,
Using gives
This is the same equation as problem 2.15 of Jackson except that the and variables are swapped. Given the fact that the problem is unchanged if the coordinates on the diagram for HW problem 5.2.1 are swapped, we are free to change replace with and vice-versa in our equation above and so we conclude that it follows from this geometrical symmetry that
Alternatively, the equivalence can be proved using an identity such as equation 3.169 of Jackson.
2. First, obtain an answer using the standard boundary value method. The potential
satisfies the three boundary conditions for which the potential on the boundary is zero.
To determine , multiply both sides by and integrate from to . This gives
and so the answer we expect is
To solve this problem using , we need to compute
(You should be able to explain what happened to the other terms in 1.42 of Jackson, which the above is based on and also the justification for the following equation.)
For this problem, the required integration is
Using and the fact that integration over gives , we have
Using the relationship
this is
Using ,
(The two other terms in involving the derivative of cancel; this was considered in a previous HW problem.) The second term in this equation is zero because the volume extends from to for which . Also, in this range of , . This leaves
Using
gives
and the integral
can be rewritten as
The integral is zero for and for . The final result is then
which is the same as the result found earlier.
1. Find for problem HW 4.1 in terms of an infinite sum involving the Legendre polynomials.
2. Find the surface charge density on the part of the dome in the – plane.
Notes
Try to work out using only the first few terms in the expansion. This will require only straightforward derivatives and the use of the table of Legendre polynomials.
Given that the exam is coming up, I’ll make getting the general equation extra credit. As a hint, use the middle equation of 3.29 of Jackson, which is to compute the derivatives of the Legendre polynomial. Given that we are always evaluating at (corresponding to ), you need to use . This will give you a general equation for the derivative evaluated at … but you’ll need to know . You can get that using the first equation of 3.29 of Jackson: with . Extra extra credit if you plot the result.
Comments
Many students attempted to solve this using reciprocity. The potential for a point charge above a grounded plane can be written in terms of the Legendre polynomials (the terms will be proportional to … one could use the azimuthal symmetry trick to show this). To use reciprocity, one needs to evaluate
where is the desired potential. The surface is the entire dome, over which is zero except for . As a result, one needs to integrate
It is possible to use this approach, but most students wrote down the wrong integral that needed to be evaluated.
Solution
1. Letting so that potential is units of , the exact solution on the –axis is
which can be written as
or
I’ve kept the absolute values because in some problems, it matters. Here it does not because is by definition positive, and we are considering the solution for . We can thus write the exact solution as either
or
Where the superscript indicates the equation will be used for either the outer ( ; ) or inner (; ) solution.
The binomial expansion has the form
For , we have
Using gives
Using this in the expansion for gives
or
with
where the double factorial is similar to the usual factorial with every other term omitted: for odd and for even.
We know that the potential for a charge distribution with azimuthal symmetry can be written in the form
This equation must match a power series expansion of the potential along the -axis, which corresponds to and .
or, equivalently,
The values of can be determined from the first recursion relationship of 3.29 of Jackson:
with , this is
The first few values can be computed manually
Based on the above, we expect that for all . (But this is not a proof.)
For this to match the expansion
we need for all and
because there is no term in the expansion.
because the constants must match.
because there is no term in the expansion.
because the constants must match.
etc.
In summary,
and, with the recovered,
We can write the coefficients more generally. They are
That is,
Shifting the index by gives
The above steps are what is needed to find the constants for the potential for . The steps for finding the potential for are similar. In this case, the terms are all zero and only every other term is non–zero.
2. From 1., you will have a potential for of the form
From Gauss’s law, the surface charge density just outside of a conductor is , where is the outward normal to the surface. Equivalently,
where the partial derivative is evaluated at the the surface.
In this problem, and we need to evaluate the partial derivative at . That is, we need to compute
Given that , we can write
To simplify notation, let
so that
For the first term, we have, using the product rule for derivatives
Given that , we are left with
At , (the radial cylindrical coordinate), so
Using the chain rule for derivatives
with and , the term in square braces can be re-written, giving
Using
leaves
Repeating the above for gives
In general,
Given that , , and so
Given that , , and so
Given that , , and so
To obtain a general solution, we need to know . This can be found using the second equation of 3.29 of Jackson:
When evaluated at , this is
Using again the first recursion relationship of 3.29 of Jackson
with , gives
so
,
,
,
,
,
or, in general
From which it follows
Shifting the index by gives
So the first few non–zero terms are
Check: Table 3.15 gives the first five Legendre polynomials. Manual evaluation of the derivatives gives:
From a table, so
In summary,
With the values found earlier, this is
Recovering the and writing in non–dimensional form gives
The leading order terms is negative as expected because the charge for should be negative given that it is grounded and the inner circle is held at a positive potential.
The general equations found earlier are
where
and
The product of the last two equations is
or
so that
and finally
(To convert this to a sum over all positive integers, define .)
In class, we started the problem of finding the potential inside and outside of a sphere of radius that is centered on the origin and is held at potential .
Find (unless otherwise stated, this type of statement means for all ).
Most students immediately re–wrote as prior to starting to answer part 1. by using the fact that and . In the general case, it will not always be obvious how to express an arbitrary boundary potential in terms of the Legendre polynomials.
In 3.2 of Jackson, he notes that an arbitrary function of (he uses , which is also ) may be written as
To find the coefficients , multiply both sides by and then integrate from to . (This is essentially what Griffiths calls “Fourier’s trick” except using Legendre polynomials). By orthogonality, one then finds that
Suppose for to and for to . Write in the form . That is, find a second–order approximation to by finding , , and . (If you use a formula from Jackson or elsewhere instead of doing integration to find and , prove the formula.)
Find using the boundary value .
State at least one way that you can determine if your answer to part 3. makes sense.
(Optional) Repeat this problem assuming that instead of being held at a potential, the sphere has a surface charge density of or for to and for to
Solution
1. By inspection of a table of Legendre Polynomials,
The general solution to Laplace’s equation is
As a result of holding the sphere at a potential, charges will build up on its surface. Far away from the surface, these surface charges will appear as a point charge. As a result, the potential should approach zero as , which requires . We then are left with
Setting , we have
“Matching terms” (meaning muliplying both sides by and integrating from to and using ) gives
Alternative (but conceptually the same) approach
In Griffiths example 3.7, the general solution for the potential outside of a sphere of radius held at is given as:
where the last step used . Plugging in the given boundary condition for gives
For ,
Using the orthogonality relationship
gives
For ,
and the orthogonality relationship gives
The term corresponds to the potential due to the net charge induced on the surface of the sphere. As a result, if we compute and integrate over , we expect to find the net charge on the sphere is such that
2. To find the coefficients , multiply both sides by and then integrate from to . (This is essentially what Griffiths calls “Fourier’s trick” except using Legendre polynomials). By orthogonality, one then finds that
Using , this is
Using for and for gives
Direct integration gives
Therefore, keeping only the first three terms of
gives
This answer makes sense – this potential is maximum at , at and minimum at . This generally follows the pattern of the exact piecewise function that we are approximating.
(Equation 3.26 of Jackson gives a general equation for , which can be derived using Legendre polynomial identities.)
3. Following steps similar to that used in part 1.
4. Given that the net charge on the sphere is expected to be zero (the potential is symmetric about ), we don’t expect a term. At , matches the boundary condition . This potential is also symmetric about , which is expected given the boundary condition has this symmetry.
An additional check that we can make is verify that the net charge is zero by computing and integrating it over .
A point charge is at
Starting with
find the potential for to order
Starting with
find the potential for to order
Show that your answers for 1. and 2. are equal
Solution
1. There are many ways of writing . In the following, I emphasis the dependence of on and by writing .
In this problem, and so , leaving
Here we have , so and and so
To order , this is
or
2. Assume that so that we do not need to carry around the . One could then repeat using (so ) to show that the same equation results. We need to evaluate the and terms of
Where was used
From a table,
Note that the potential must not be imaginary, so the algebra must be such that is not in the solution. Here we have , so , leaving
Substitution of equation from the table gives
Using and gives
Using the Euler identity for complex exponentials gives
for the term. The sum of the and terms are thus
This could also be written in cartesian coordinates as
3. Not applicable b/c the answer to 2. was simplified so it matches by inspection.
Checks:
If , we get , as expected.
Symmetry with respect to and .
A straight line of charge with uniform charge density extends from to .
Starting with
find the potential for to order
Starting with
find the potential for to order
Use either 1. or 2. to find for to order .
Solution
1. Assuming again that so we do not need to carry around the ,
as in the point charge case (all differential charges are on the –axis) and so it can be factored out of the integral (in general, this term cannot be factored out, however; recall that in general ). Also, does not depend on , so it can be factored out of the integral. Thus,
Note that the 2nd, 4th, … integrals are zero because their integrands are an odd function on the integration interval. This leaves
or
or, using
2.
Factoring out all terms that do not depend on primed variables gives
and do not depend on because for all differential charges on the line, so can also be factored out
Using the algebra from the point charge expressed using spherical harmonics will given the same result as in 1.
3.
Using and gives, for the first term,
The net charge on the line is and so this leading order term is as expected (far from the line, the potential should be , where is the charge on the line). There is no “dipole” term () because the charge distribution is symmetric about the origin (see 3.4.2 of Griffiths, which give an alternative equation for the dipole term in terms of the dipole moment expressed as an integral).
A cylinder of radius and height is centered on the origin, has a centerline aligned with the –axis, and has a “frozen–in” polarization of .
Compute and
Find the potential on on the –axis for
Find the potential to order for .
Solution
1. The bound charge associated with with polarization is on the top/bottom caps of the cylinder and . As a result, the electric field produced by this is the same as the electric field produced by two disks with for which several methods are available for computing .
2. The potential for a disk in the – plane and centered on the origin is
(the derivation is treated in intro physics textbooks). As a check, for , the leading term is and for , the first–order term in the electric field is (the potential in this limit is ).
The potential for the given case is found by replacing with for the top cap and for the bottom cap and using for the top cap and for the bottom cap, so that
3. This problem has azimuthal symmetry and so can be found for small by first expresing as a power series expansion in . Factoring out gives
where
and .
There is a significant amound of algebra involed in obtaining a power series expansion. WolframAlpha gives
For , this is
and so to order ,
One issue not addressed is for what (or ) this applies to. The expansion of requires , corresponding to and . However, I asked for a solution valid for , which is less than . The solution above actually valid for even though this does not follow from the arguments made for deriving it. A similar issue arises in 3.4.1 of Griffiths where he derives the monopole expansion. In that derivation the parameter must be less than one and the resulting equation is actually valid for even though the , where can vary between and .
One argument that can be made for the validity of the solution given above for instead of only is as follows. We know for a charge at , the solution can be expressed by an expansion of as a sum of terms, where is a constant, which is valid for . In the given problem, all of the charges are at and so we expect that such a solution exists. However, we found a solution in this form, and by uniqueness, it must be the only solution.
An alternative way of showing this is to use the equation for the potential of a ring centered on the origin and offset from the – plane, which is given on page 104 of Jackson:
where, using the variables in this problem, and . This equation is valid for (that is, for ), which was asked for in the problem statement. To find the potential for the positively charged disk, we need to integrate rings. This is done by replacing with and in and with . With this, we need to evaluate
When this is added to the potential for the negatively charged disk, the terms cancel leaving
For both the and terms, the integral is straightforward and the result is the same found earlier.
A point charge is at the origin. A thick dielectric shell with susceptibility with inner radius and outer radius is centered on the origin.
Find using at least two different methods.
Solution
Approach 1.
Gauss’ law for dielectrics give in all three regions and symmetry requires . Using gives
Check: As , the electric field in the dielectric is zero and the other electric fields are as expected from Gauss’ law. As , the field in all regions is that for a point charge.
Using and can be used with the for to find for .
Then can be used to find for , where is known from the potential found for .
The potnetial for can be found in a similar way. The result is
The previous two checks apply. In addition, we expect continuity to apply, so the potentials should match at their boundaries.
Approach 2.
Write in all three regions. There will be six unknowns. The six conditions are , , continuity at and , and the jump condition at and .
Approach 3.
This approach is similar to the one used in class to introduce polarization. Assume that inside the dielectric , where is the field due to bound charges. For a linear dielectric, . We can use to find the and and then use Gauss’ law with , where is the bound charge enclosed.
The enclosed bound charge for the surface of a Gaussian sphere is
where
and
The volume integral could also have been computed using the divergence theorem:
Finally, Gauss’ law
from above and gives the same electric field inside of the dielectric as found earlier. The electric field in the other regions can be found using the same steps. You should find the enclosed is zero inside of a Gaussian sphere of and for , because the dielectric is not inside of the Gaussian sphere.
If you compute after finding (and hence is also known), you will find .
A conducting shell of radius has a potential of , where is the potential found in part 1. A conducting shell of radius is at potential , where is the potential found in part 1. A thick dielectric shell with susceptibility with inner radius and outer radius is between the two conductors.
Starting with and , use the boundary conditions, continuity, and the jump condition to find the potential in the region between the two conductors.
Use your answer to 1. to find the electric field between the two conductors.
Solution
From uniqueness, you should end up with the same result as the previous problem.
A long solid cylinder of radius is aligned with the –axis and centered on the origin. The rod is polarized such that .
Find
due to the bound surface charges only. Assume the cylinder has .
Solution
1. See HW 5.3. Steps are similar.
Note that the bound charge is
To find the contribution from the bound charge, one could find the potentials and due to a shell with density . The potential at is then the sum of the potentials due to differential shells from to and to .
A perfect dipole at the origin is at the center of a dielectric sphere of radius that is centered on the origin. The dielectric constant of the sphere is .
Find
and
Explicitly compute for a surface of an origin–centered sphere of radius .
Solution
1.
Checks: When , we have simply a dipole in free space and .
If is large (so is large), the dielectric becomes like a conductor. In this case, the electric field can be made to be zero if positive charges cluster in shell of vanishing radius around the negative charges in the dipole and if negative charges cluster around the positive charge in the dipole. In the limit that is large, , which is consistent with this.
It would be useful to compute to verify that it is consistent with the stated distribution of charges for large . There are two ways to do this: (1) Compute . We expect that should be proportional to in the limit that . Showing this takes a bit of delta function manupulation. Alternatively, (2), we could re–do the problem using point charges at . In this case, we will have more terms in , but will be able to more clearly identify and when computing .
An origin–centered dielectric sphere with radius has a susceptibility of . A point charge is at , where .
Find the potential. Note that as , the solution should match that of a point charge outside of a conducting sphere.
Hints:
The potential will be due to the point charge and the bound charges induced on the sphere. You will need to consider three regions:
Inner
Middle
Outer
In the outer region, the potential due to the point charge can be expressed (from 3.38 of Jackson) as
In the inner and middle region, the potential due to the point charge can be expressed as
In all three regions, the contribution from the induced bound charges must have the form
In the inner region, the potential due to the bound charges will not have terms. In the middle and outer region, the potential due to the bound charges will not have terms.
Solution
One could start with
for the inner (), middle (), and outer () regions and the conditions
where . In this case, we need to find , , , and .
Using the hint, we one start with
Where now we need to find , , and . In the middle region and outer region, the contribution from the dielectric must be the same and so we can conclude .
Therefore, we only need to find and in
and can be found using and . The arguments provided obviate the need to use the boundary conditions at because they were implicitly used in simplifying the set of equations to be solved.
The circular loop with radius lies in the plane () and the axis passes through its center. There is a current in the loop in the direction shown in the following figure.
A magnetic dipole at the origin creates a magnetic field of
1. Use eqn 5.12 in Jackson 3rd ed. () to find the force on the loop in cartesian coordinates.
2. Find the force on the loop in cartesian coordinates using eqn 5.69 of Jackson 3rd ed. ().
3. Write your answer to 1. as a power series in terms of powers of (assume .
A hollow and open-ended cylinder of height and radius is centered on the origin and aligned with the -axis. The curved surface of the cylinder has a surface current of .
1. Find using the Biot–Savart formula
2. Use the formula
to express as a power series involving the Legendre polynomials, , and .
Read 5.6–5.12 of Jackson and Chapter 6 of Griffiths and be prepared to ask question during class. (Questions on Discord are also welcome).